Question

A 17.0 g bullet traveling horizontally at 785 m/s passes through a tank containing 13.5 kg of water and emerges with a speed of 534 m/s.

What is the maximum temperature increase that the water could have as a result of this event? (in degrees)

Answer 1

The maximum temperature increase is
`\Delta T = 0.0497 \ ^oC`

Step-by-step explanation:

From the question we are told that

The mass of the bullet is
`m = 17.0 \ g =0.017 \ kg`

The speed is
`v_1 = 785 \ m/s`

The mass of the water is
`m_w = 13.5 \ kg`

The velocity it emerged with is
`v_2 = 534 \ m/s`

Generally due to the fact that energy can nether be created nor destroyed but transferred from one form to another then

the change in kinetic energy of the bullet = the heat gained by the water

So

The change in kinetic energy of the water is

`\Delta KE = (1)/(2) m (v_1^2 - v_2 ^2 )`

substituting values

`\Delta KE =0.5 * 0.017 * (( 785)^2 - (534) ^2 )`

`\Delta KE = 2814.1 \ J`

Now the heat gained by the water is

`Q = m_w* c_w * \Delta T`

Here
`c_w` is the specific heat of water which has a value
`c_w = 4190 J/kg \cdot K`

So since
`\Delta KE = Q`

we have that

`2814.1 = 13.5 * 4190 * \Delta T`

`\Delta T = 0.0497 \ ^oC`